Momentum gauge invariant

For the Maxwell field, the energy-momentum tensor Tfi(A) derived from Noether s theorem is unsymmetric, and not gauge invariant, in contrast to the symmetric stress tensor derived directly from Maxwell s equations [318], Consider the symmetric tensor 0 = T + AT, where... 

The S U(2) energy-momentum tensor can be symmetrized and made gauge invariant by adding an incremental tensor... 

If the physical results are to remain unchanged under a unitary transformation, it is necessary to transform the operators as well as the state functions (eqns (8.28) and (8.29)). A gauge transformation has no effect on a coordinate operator but the momentum operator p is changed into p + [e/c) x- Thus, to maintain gauge invariance in properties determined by the momentum operator, p must be replaced by a new operator. In the presence of an electromagnetic field or just a magnetic field, the momentum operator is replaced by the expression... 

This expression of K is gauge-invariant, and both of the first and the second terms on the right-hand side of Eq. (10) are independent of the choice of body frame. The former vanishes if and only if the total angular momentum L is zero [cf. Eq. (5)]. The metric tensor g in Eq. (11) is also gauge-invariant and is the true metric appropriate for the description of internal motions in polyatomic molecules. 

Note that we have introduced the symbol 7t for linear momentum here in order to better distinguish it from canonical momentum. This notational rigor is only needed for the discussion of this section and will thus be dropped elsewhere in the book. In most cases it will become obvious from the context to which kind of momentum we are referring. Obviously, it is rather the linear than the canonical momentum which is gauge invariant. The canonical momentum p satisfies the fundamental Poisson brackets of Eq. (2.81), of course, whereas the components of linear momentum, interpreted as phase space functions tt, = Ki r,p,t), feature nonvanishing but gauge invariant Poisson brackets. 

The form invariance of the Schrodinger equation will then lead to gauge invariant expectation values of the Hamiltonian. However, this will not be the case for an arbitrary operator. In particular, it turns out that expectation values of the canonical momentum operator, given in Eq. (2.45), are not gauge invariant, whereas expectation values of the mechanical or kinematical momentum operator, given in Eq. (2.97), are gauge invariant [see Exercise 2.15]... 

The mechanical or kinematical momentum operator is therefore sometimes also called the gauge invariant momentum operator. 

The more general form of (5.10.2) is obtained on introducing field terms in the Hamiltonian via the gauge-invariant momentum operator , whose components (see Section 11.1) are... 

Finally we must verify that the energies and densities obtained from H are indeed gauge-invariant. To do this, we first show that the gauge-invariant momentum operator (taking first a one-electron system)... 

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